Contents

Lessons 21 - 30 2Notes 21-22 – Translating Lines, Circles and Squares . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Assignment 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Assignment 22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Notes 23 – Stretching and Squishing Lines, Circles and Squares . . . . . . . . . . . . . . . . . . . . 8Assignment 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

Notes 24 – Slope-Intercept and Standard Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13Assignment 24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

Notes 25-26 – Getting Faster at Writing Equations of Lines . . . . . . . . . . . . . . . . . . . . . . 17Assignment 25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20Assignment 26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

Notes 27 – Solving Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22Assignment 27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

Notes 28 – Linear Story Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28Assignment 28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

Notes 29 – Angles and Parts of Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32Assignment 29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

Notes 30 – Angles and Slopes - Part I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34Assignment 30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Please note that the decision to include exactly ten assignment (along with their relevant typed up set ofnotes) was driven by file size, not by unit and/or chapter length. I will be including a file with assessments(along with indications of what assignments they cover) on my webpage shortly.

9th Honors Math Notes for 19th and 20th Assignments

Translated Circles and Squares (and Lines) First, an observation. Look at these two equations: 5(x) = 20 5( x – 3 ) = 20

Notice that the answer to the first equation is x = 4 and the answer to the second is x = 7. The answer got LARGER by 3, when the x was replaced by x made SMALLER by 3.

Can you see that replacing x by something LARGER than x would cause the answer to get SMALLER? Let’s play these ideas with the graphs of squares and circles.

Old Business New Business The graph of makes The equation looks a square. How did we know? a lot like the previous equation, Find some ordered pairs and but the points that work in it are graph them. See the ordered different, in an interesting and pairs at right and picture below? predictable way. Here is a table The highest point is ( 0 , 4 ) and for it. See that the numbers for the lowest point is ( 0 , -4 ). the second coordinate, y, are each Its center is ( 0 , 0 ) one larger than they were before? That means that graph will be one higher!! The center is ( 0 , 1 ). Why does this happen? We can put in larger numbers for y because we subtract 1 from that number before we do the rest of the arithmetic. Kind of a compensation thing. It’s called “translating”/migrating the shape. How might we write an equation so that the graph would have gone three lower instead of one higher?

Answer: Write for lowering it three instead of for raising it one.

Are you noticing that adding to the variable, makes you subtract from the coordinates, so that the effect is opposite what you might initially be expecting? One thing that should seem reasonable, is that if you affect the y coordinate, that the points will then move in a y (up/down) direction. Here are equations for two circles The first has its center at ( 0 , 0 ) The second has its center at ( 2 , 0 ). At right are the two tables and The graphs of each circle.

x2 + y2 = 25 ( x – 2 )2 + y2 = 25 This time we altered the x-coordinates, and the shaped moved sideways ( in the x-coordinate direction).

x y 4 0 3 1 2 2 1 3 0 4 0 -4

x y 4 1 3 2 2 3 1 4 0 5 0 -3

x y 5 0 4 3 3 4 0 5 0 -5 -5 0

x y 7 0 6 3 5 4 2 5 2 -5 -3 0

2

Can you guess the effect of changing into ? We’ve made two replacements. We replaced x with x – 5 .......which will migrate every point 5 to the RIGHT. We replaced y with y + 3....... which will migrate every point 3 to the DOWN.

It’s still a square, but the center of the square is now ( 5 , -3 ). The highest point will be 4 higher than the center, at ( 5 , 1) and the rightmost point will be 4 to the right of the center, at ( 9 , -3). Check them in the equation. The cool thing about migrating the points is…AREAS and PERIMETERS DON’T CHANGE!!!! Can you guess the effect of changing into ? We replaced x with x + 6......... which will migrate every point 6 to the LEFT. We replaced y with y – 4......... which will migrate every point 4 to the UP. It’s still a circle, but the center of the circle is now ( -6 , 4 ). Since the radius of the circle is 7, the highest point of the circle will be 7 higher than the center, at ( -6 , 11 ), and the leftmost point will be 7 to the left of the center, at ( -13 , 4 ). AND…..AREAS and PERIMETERS DON’T CHANGE when you migrate!!!! Here are some equations and their graphs. Could you have graphed them if you had been given the equations, and could you have given the equations if you had been given the graphs?

( x – 4 )2 + (y-1)2 = 9

x2 + (y-4)2 = 4 We say that the two circles are tangent to each other. That means that they just BARELY touch at one point. It also means that the point of tangency is collinear with the two centers. How do we know this, since it’s hard to read the coordinates of that point? We know because the distance between ( 0 , 4 ) and ( 4 , 1 ), the centers of the circles, is 5. Also, the radii of the circles are 2 and 3, whose sum is 5. That means that there is ONE point that is on the segment connecting their centers, and that is on the circumference of each circle, a point common to each circle. Lines translate/migrate in a similar fashion. The line y = ½ x passes through the origin. The line y - 2 = ½ ( x – 1) doesn’t. However it’s just the previous line, translated so that the point that WAS at the origin is now at ( 1 , 2 ). Here’s the picture. It’s just the line passing through ( 1 , 2 ) , with a slope of ½ . See the arrow, showing how the origin migrated? This style of writing lines is called point-slope…because when you look at a line in this form ( y – k ) = m( x – h ), you can graph it quickly, through the point ( h , k ) with a slope of m.

3

Ninth Honors Name _____________________________ 21st Assignment Period _______

1. What is the slope of the segment between

!

12,1.375

"

# $

%

& ' and

!

0.125," 34

#

$ %

&

' ( ? 1. ________

2. Find the distance between the points

!

("0.36,0.7) and

!

4 711,3.4

"

# $

%

& ' . 2. ________

3. Draw the graph of each equation listed below. a)

!

x + y = 4 b)

!

x 2 + y + 3( )2 = 9 c)

!

x +1 = 3 " y " 2 d)

!

y " 2 = "0.25(x + 3) 4. Write an equation for a circle whose circumference is 1/2 as much as the one for 4. ________ 5. The perimeter of the square whose equation is

!

x + y = 3 is closest to what whole number? 5. ________ 6. Of all the points on the graph of

!

x " 60( )2 + y + 40( )2 = 441, which point is closest to the y-axis? 6. _______ 7. Solve each equation below for x. Give every answer, in reduced form. 7. a) ______

a)

!

x2

+13

=1118

x "29

b)

!

x 2

3"12

=x 2

4+56

c)

!

2x " 5( ) 25x 2 " 9( ) = 0 b) _______

c) _______

d)

!

0.27___x =1.54

___ e)

!

5(x + 4) " 6(x " 5) = 2(x " 6) d) _______ e) _______ 8. Convert each decimal into a fraction (ratio of integers) in lowest terms. 8. a) ____ b) ____

a) .72 b)

!

1.285714 c)

!

.133___

d)

!

0.833___" 0.083

c) ____ d) ____ 9. What is the area of the square whose equation is

!

x " 2 + y + 3 = 9 ? 9. _______ 10. Write an equation for a line passing through (0.3,-0.5) with a slope of 2/13 10. ____________________ 11. A line passes perpendicularly through the midpoint of the segment connecting A(4,2) 11. ________ and B(10,0). What are the coordinates of the point where the line crosses the x-axis?

12. For the circle whose diameter connects

!

- 112,1 34

"

# $

%

& ' and

!

1 512,- 14

"

# $

%

& ' ... 12. a) ________

a) what is its exact area? b) what is the approximate distance its center is from the origin to the nearest hundredth? b) ________

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9th Honors Name _____________________________ 22nd Assignment (graph paper is on back…) Period _______ 1. By how much does the area of exceed the area of ? _____ 2. A square has an area of 8x + 9. When the perimeter of the square is 28, what is the value of x? ______ 3. If a circle whose center is ( 2 , 3 ) has the point ( 6 , 6 ) on its circumference,

a. what is the radius of the circle? ________ b. what are the coordinates of the highest point on the circle? _________ c. what is the exact circumference of the circle? _______ d. what are the coordinates of the point on the circle opposite (6 , 6 )? ________ e. what is an equation of the circle? ________________________

4. When x=64 and y=75, how far apart are the points ( x + 3 , y + 5 ) and ( x + 8 , y + 10 ) ? _______ 5. A line passes perpendicularly through the midpoint of the segment A(2 , 8 ) B( 4 , 0 ). What are the coordinates of the point where the line crosses the x-axis? ______ 6. Find a point on y – 5 = – ! ( x + 2 ) with natural number coordinates. ___________ 7. For the circle whose equation is

!

x + 3( )2 + y " 4( )2 = 25, what is the length of that part of the circle’s circumference that is in the second quadrant? ______ 8. At what two points does hit the y-axis? _______ and ________ 9. For a certain value of x, both 5x +4 and 7x – 20 represent the area of a square. For that x, what is the perimeter of that square? _______ 10. What’s the area of ? _______

11. How far is the top of

!

x " 3( )2 + y +1( )2 = 9 from the top of ? _______

12. Graph each group of equations on the accompanying axes

a) b) c) d)

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Notes for the 23rd Assignment Stretching and Squishing Squares and Circles Here is a piano for you… Here’s how you play it. Pick some numbers, some positive some negative, some not integers, and mentally place your fingers on those spots. Then listen to the musical instructions from your teacher and move your fingers so as to “play that chord”. For example, suppose the chord is the “+3” chord. That means add three to every number you previously had your fingers on, and essentially move everything to the right 3. Consider the effect of each of the following chords listed below, and see if you can predict how your fingers would move in each case. “ subtract 4” “subtract –0.5” “multiply by 2” “ multiply by ¼ ” “ multiply by -1” “divide by 3” “ multiply by – ½ ” “divide by 1/3” “multiply by zero” “reciprocal” “absolute value” Did you notice how multiplying and dividing each number produced either a stretching apart or a squishing together? We can exploit that with our equations of squares and circles.

Old Business New Business

The graph of is The graph of

(still) a square. How do we know? is similar, but the points that work in it Either we have that type of graph are different and need to be explored. memorized, or we find some Here is a table of some points for it. some points and graph them. Notice that the first coordinates are The highest point is ( 0 , 3 ) and twice as much as they used to be. The the lowest point is ( 0 , -3) while and the highest point is still ( 0 , 3 ) but right and left most points are ( 3 , 0 ) and ( -3 , 0 ) the right and left most points become (6, 0 ) & (-6,0) Why does this happen? We can put in larger numbers for x because we cut them in half before we do the rest of the arithmetic. It’s kind of a compensation thing, like the migrating was, but this time involving multiplying and dividing. Can you see how we might squish it vertically instead of stretch it? Yep, an equation like

would make no number larger than 1.5 eligible for the variable y.

x y 6 0 4 1 2 2 0 3 0 -3 -6 0

x y 3 0 2 1 1 2 0 3 0 -3 -3 0

8

How might we write an equation so that the graph would have been stretched to three times its width?

Answer: Write to allow for numbers as large as 9 to be placed in for the variable x.

So, what do you think happens to the area and the perimeter when we stretch everything twice as high? You’re right and you’re wrong. If you said, “they will likely also change”…you’re right If you said, “they will likely become twice as much”, you’re right and you’re wrong. If you said, “the AREA will become twice as much”, you’re right. Check it out by looking at the two pictures. In the first case, the area is 18, while in the second case, the area is 36. However, notice that the perimeter changes in a peculiar way. The sides are no longer each; they are now

each, so the entire perimeter is now instead of . Strange, but do-able…for squares that become stretched into rhombi that are no longer squares Circles, ah, that’s another story. Here is a graph of the circle whose equation is and some points to help us see the curving. Notice that the radius is 3, so the area is 9π and the perimeter/circumference is 6π.

Now, here is a graph of the “stretched circle”, really called an ellipse, whose equation is

x y 3 0

0 3 0 -3 -3 0

x y 6 0 2

0 3 0 -3 -3 0

9

The area and perimeter of ellipses is harder to determine than for rhombi…OK, half harder. Can you guess which half is harder? Think about which half was harder for squares and rhombi. Areas behave just like when we stretched squares to rhombi. Twice as wide, twice the area, so this ellipse has an area of twice 9π, or 18π. Perimeters of ellipses are HARD to figure out, so hard that you’ll have to wait until you are second semester seniors before we can adequately handle that one.

Ok, one last observation, concerning the ellipse whose equation was .

Do you see that the first thing squared is a fraction? It’s , and since squaring means to multiply by itself,

that you must have , and that equals . Think about it…that is a synonym for , but it’s harder to

look at and know that the ellipse got stretched to twice its width, unless you think of it as and realize

that EACH x was stretched to twice its width. It’s just September 29th…multiplying fractions.

So, here’s a second example of ellipses: . Its top point (when x=0) is at (0 , 5 ). Its bottom

point is at ( 0 , -5 ). It’s rightmost point is ( 20 , 0 ) because it was stretched 4 times as wide, and also simply because when you put 20 in for x, and then divide by 4, you get 5, which is the perfect number to square if you want to get 25.

Here’s a third example of ellipses: . Do you see that it’s the previous example, multiplied out?

Ok, a mixture of stretched and squished stuff.

….circle with radius of 5, but then stretched 3 times as wide.

…circle with radius of 6, but then squished to half as wide.

…circle with radius of 6, but then squished to half as wide and a third as high.

…same as the previous example, with the 2x times 2x and 3y times 3y multiplied out.

If this is confusing, just plot some points…you’ll see that when x=0, y must equal 2 or –2 , and when y=0, x must equal either 3 or –3. Then you know it’s a morphed circle, so just ellipse-ize it.

…a square whose corners used to be all 8 away from the origin, but whose high and low points are now only half has high from the origin. It’s a rhombus whose main/major diagonal has length 16, and whose minor diagonal only has length of 8.

…the square from above, but stretched three times as wide and squished to half as high. 10

9th Honors Name _____________________________ 23rd Assignment Period _______

Circle the word in each pair that makes the sentence correct.

1. The graph of was a square/circle that was stretched/squished horizontally/vertically.

2. The graph of was a square/circle that was stretched/squished horizontally/vertically.

3. The graph of was a square/circle that was stretched/squished horizontally/vertically.

4. The graph of was a square/circle that was stretched/squished horizontally/vertically.

5. When the variable x is replaced by 4x, what happens to the graph? ______________________________

6. When the variable y is replaced by , what happens to the graph? ______________________________

7. Graph on the grid at right.

a) what is its area? ______ b) what is its perimeter? ______

8. Graph on the same grid at right.

a) what is its area? _______ b) what is the length of its shortest/minor diameter? ______ c) what is the length of its longest/major diameter? ______

9. For the ellipse shown at right, what is …

a) its exact area? ______ 6 b) the length of its major diameter? ____ c) the length of its minor diameter? ____ d) an equation? 3

______________________

10. For the rhombus shown at right, what is … -6 -1 1 6

a) its exact area? ______ b) the length of its major diagonal? ____ c) the length of its minor diagonal? ____ -3 d) its exact perimeter? ______ e) its approximate perimeter? _______ f) an equation? -6

______________________

11. For the ellipse whose equation is and the rhombus whose equation is ,

how far is the top most point of the ellipse from the rightmost point of the rhombus? _______

11

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Notes for the 24th Assignment

Slope-Intercept and Standard Forms Two Ways to Write Equations of Lines

When a linear equation is written in this form, 4x + 3y = 36 , with the x and y variables on one side and the constant term on the other side, it is said to be written in Standard Form.

When a linear equation is written in this form, 1234

+-= xy , with the y “solved for” on one side of

the equation and everything else on the other side, it is said to be written in Slope-Intercept Form. The two equations above represent the same line, written in the two styles. Let’s explore how easily one can glean the important information about the line from each style. y-intercept: Recalling that the y-intercept is the point on the y-axis, we know that its x-coordinate is zero. We can put zero into the Standard Form and get 3y = 36, from which we see that the point is ( 0 , 12 ). We can also see that point immediately from slope-intercept form because when x=0 , it boils down to simply y = 12. x-intercept: Recalling that the x-intercept is the point on the x-axis, we know that its Y-coordinate is zero. we can get that just as easily from Standard Form, and get 4x = 36, from which we see that the point is ( 9 , 0 ).

We can get that from slope-intercept form as well, but not as easily. We put 0 in for y and get 12340 +-= x .

Now we have to solve the equation for x, by subtracting 12 from each side to get x3412 -=- and then

multiplying each side by 43

- , which also yields x=9, just more work.

Slope: One of the selling points of slope-intercept form is that the slope may be immediately seen. It’s the

coefficient of x, which is 34

- . That piece of information about the line may also be quickly seen when the

equation is written in standard form…notice that 4 and 3 are the coefficients of x and y. When the equation is

written in the style Ax + By = C, the slope is just BA

- , also pretty easy to make use of.

Now there is no need for you to have to choose a favorite candidate, like the presidential election. You can like BOTH styles, and use whichever is easiest for the situation you’re given. Some examples:

Write an equation for a line whose slope is 31

- and whose y-intercept is 8.

1. Slope Intercept: Simple, just fill in the blanks. 831

+-= xy

2. Standard Form: Almost as simple. Make the left side… =+ yx 3 ? (so the slope is 31

- ) and

then put in 24 for the question mark. Why 24? Because you want the y-intercept to be 8, so you want ( 0, 8) to work.

13

Write an equation for a line whose slope is 32 , and whose x-intercept is 5.

1. Slope Intercept: Not as simple… fill in the blanks. ?32

+= xy , and put in 310

- for the

question mark because you want the point ( 0 , 5 ) to work. How do you decide 310

- ? You

have to solve the equation ?5320 +×=

2. Standard Form: Exactly as simple. Make the left side ?32 =- yx and put in 10 for the question mark because you want the point ( 5, 0 ) to work.

Write an equation for a line whose slope is 135

- , passing through the point ( 4 , 1 ).

1. Slope Intercept: Very not as simple… fill in the blanks. ?135

+-= xy , and put in

…WHAT…for the question mark because you want the point ( 4 , 1 ) to work. How do

you decide what works? You have to solve the equation ?41351 +×-= which leads to

to a y-intercept of Slope Intercept: Not as simple… fill in the blanks. ?32

+= xy , and

put in 310

- for the question mark because you want the point ( 0 , 5 ) to work. How do

you decide 310

- ? You have to solve the equation 1333 . Not fun.

2. Standard Form: Exactly as simple. Make the left side ?135 =+ yx and put in 33 for the question mark because you want the point ( 4, 1 ) to work. Done.

Hmm, there’s merit in this Standard Form. Same approach, same level of difficulty, very few fractions.

Let’s practice a few things with Standard Form. Tell me the intercepts and slope for each line described below.

3x + 4y = 24. ( 0 , 6 ) ( 8 , 0 ) and slope = 43

-

2x – 5y = 40 ( 0 , -8 ) ( 20 , 0 ) and slope = 52

0x + 3y = 12 ( 0 , 4 ) and HMMM and slope = 30

- , which is just zero.

Oh yeah, it’s just the equation y = 4, which is horizontal, and doesn’t have an x-intercept.

Write an equation in Standard Form for the line with slope 75

- , that goes through ( 1 , 2 )?

Actually you can, and quickly. The left side of the equation must look like this

5x + 7 y in order to have a slope of 75

-

and the right side must be the kind of number that makes ( 1 , 2 ) work…so we have 5x + 7y = C 5(1) + 7(2)=C from which we see that C=19 Hence, our equation is simply 5x + 7y = 19

14

If you WANT slope-intercept form…it’s easy to create…just solve for y, with those same 2 little steps. 1. subtract your x term to the other side, to get 7y = -5x + 19

2. divide each side (each TERM of each side) by 7, to get 719

75

+-= xy

Notice that the slope is 75

- , just like we needed, and that the y-intercept is 719 , just like “plug in zero” said.

For those of you who are “fraction nervous” , you may wish to consider taking the equations that have fractions in them, and converting them to standard form before you explore the intercept stuff.

For example: if you take 719

75

+-= xy and multiply each side by 7, to get back to 7y = -5x + 19

and back to 5x + 7y = 19, the x-intercept is obvious, (plug in zero)…it’s ÷øö

çèæ 0,519

.

Two last thoughts….What’s an equation for a line that goes through the point ( 3 , -4 ), with the line parallel to the line whose equation is 7x + 3y = 98? The 98 part might look scary, but it’s not… the two lines are parallel…so their left sides must both look the same…like this. 7x + 3y You just need a value of C so that ( 3 , -4 ) works…plug it in 7(3) + 3( -4 ) = 9 Hence the equation is 7x + 3y = 9 Done. Fini. That’s all it takes. Very last thought… What’s an equation for a line that goes through the point ( 3 , -4 ), with the line perpendicular to

the line whose equation is 7x + 3y = 98? I love doing this in Standard Form. Can you see that the left side of the equation will just be 3x – 7y = That’s how easy perpendicular can happen in standard form. Right side? No problem, you want ( 3 , -4 ) to work, so plug it in. 3(3) – 7(-4) = 9+28=37, so 3x – 7y = 37 Sweet.

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9th Honors Name _______________________________ 24th Assignment Period _____

1. Circle every equation below whose graph has a slope of 53

-

3x – 5y = 17 3x + 5y = 22 7553

+-

= xy xy53

= 53

-=xy 5x + 3y=13 6x + 10y=1

2. Circle every equation below whose graph has a y-intercept of ( 0 , 6 ).

y=3x+6 xy536 += 3x + 2y = 6 6x + 6y = 6 2x + y = 12 2x + 3y = 18 2x - 3y = 18

3. Circle every equation below whose graph has an x-intercept of ( 4 , 0 ).

y=3x + 4 xy += 4 3x + 2y = 4 x + y = 4 2x + y = 8 4x + 4y = 4 ( )4563

--

= xy

4. Graph each equation below

a) 1232 =+ yx b) 63 =- yx c) 3=+ yx

5. Fill in the table at right. Standard Form equation slope y-intercept x-intercept

6. What is the x-intercept of a line whose slope is 43

- and whose y-intercept is ( 0 , 5 ) ? ______

7. What is the y-intercept of a line whose slope is 45 that passes through the point ( 2 , 9 ) ? ______

8. What is the y-intercept of a line that passes through the points ( 2 , 5 ) and ( 7 , 11 ) ? _______

3035 =+ yx

1032 =- yx

43

- ( 0 , 5 )

( 0 , 2 ) ( 4 , 0 )

21

( 7 , 0 )

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Notes for the 25th and 26th Assignments

Writing Equations of Lines We’re going to play a game. The game is called, “What’s NOT My Line?” Here’s how it works. I’m thinking of a line. I’ll give you the opportunity to guess, by elimination, its equation, from four choices. You guess a choice that you know ISN’T the line. You have all of the tools you need. Good luck.

First Line: It has a slope of 21 and it goes through the point ( 4 , 7 )

Your choices are

A) 221

+= xy B) 12 -= xy C) 521

+= xy D) 11=+ yx

What do you say? You say, “B and D are out because the slope of B is 2 and the slope of D is –1.

Then you say, “A is out because the point ( 4 , 7 ) doesn’t work. since 24217 +×¹

Then you craftily say, “Perhaps that teacher is trying to trick me. Rather than guess C because of the process

of elimination, I’d better check to see if ( 4 , 7 ) works in it….and 54217 +×= , so it IS C !!

Second Line: It has a y-intercept of ( 0 , 3 ) and an x-intercept of ( 5 , 0 ). Your choices are all Standard Form this time: A) 3035 =+ yx B) 1535 =- yx C) 1553 =+ yx D) 3053 =+ yx What do you say? You say, “Hmmm, I can find intercepts, by just letting one variable become zero.” Quickly you discover that A is out because 303305 ¹×+× Just as quickly, you discover that B is out because 150355 ¹×-× , and that D is out because 303503 ¹×+× You carefully check choice C, and observe that ( 0 , 3 ) works because 153503 =×+× and that ( 5 , 0 ) also works because 150553 =×+× . Since there aren’t two different lines that both contain those points, C it is!!! Third Line: It has a y-intercept of ( 0 , 5 ) and an x-intercept of ( 4 , 0 ) Your choices are

A) 554

+-= xy B) 754

+= xy C) 545

+= xy D) 545

+-= xy

What do you say? You say, “All of the equations shown give a y-intercept of 5 except for B, so it’s out. Then you say, “The slope connecting ( 0 , 5 ) and ( 4 , 0 ) is negative, so C is also out. Then you say, “To get from ( 0 , 5 ) to ( 4 , 0 ), you must rise 5 and run 4, so the slope is

45

- . So you conclude, “D has the appropriate slope and the

appropriate y-intercept. Let’s just check to see if ( 4 , 0 ) works….and 54450 +×-= , so it does, and it’s D!!

17

Fourth Line: It goes through the points ( 3 , 7 ) and ( 5 , 10) Your choices are……….up to you. You find it, without the help of choosing from among four. What do you say, after you’re done complaining about me not giving you A,B,C, and D? You say, “Hmm, let’s find the slope. It’s rising 3, from 7 to 10, and running 2, from 3 to 5. That means the slope is

23 . That means the equation, in slope-intercept form, looks like this ?

23

+= xy How to find the y-

intercept? Lots of ways. Since it goes through the point ( 3 , 7 ) you could back up and find the coordinates for ( 2 , ? ) ( 1 , ? ) and then ( 0 , ? ). That will work. Those coordinates are ( 2 ,

215 ) ( 1 , 4 ) and ( 0 ,

212 ).

Therefore, the equation is 212

23

+= xy . That’s one way.

Or, you could call the y-intercept ( 0 , B ) and put those into the equation with ( 3 , 7 ) like this: B+×= 3237

Then you could solve the equation for B, by subtracting 214 ( which is 3

23× ) from each side. Look at that!!

You discover that B=212 , so you again know that the equation is

212

23

+= xy .

Or, you fraction-phobics could say, “If the slope is 23 , there’s a good chance the equation will be riddled with

fractions. Let’s see how to avoid that. Hmmm, I bet I could write it like this: =- yx 23 something, because if you isolate the y by solving for it, you’ll get that slope. I wonder what number goes on the right side, where the “something” is? Oh wait…I remember that from yesterday… “The points WORK in the equation. That means you can put in ( 3 , 7 ), which makes the left side look like this: =×-× 7233 and that makes –5. Aha!! Another way, perfectly legal and fractionless, is to write the equation as 523 -=- yx Fifth Line: It goes through the origin, and is perpendicular to 3x + 7y = 19 Your choices are A) 073 =+ yx B) 037 =- yx C) 037 =+ yx D) 1037 =+ yx What do you say? You say, “All of the equations shown go through the origin except D, so it’s out. Then you say, “The slope of A is identical to the slope of the line you gave us, so they are parallel, and just because it’s a long word that starts with the letter P, that’s not good enough. A is out.” Then you say, “Aha, the slope of B is

37 and it

goes through the origin. It’s B !!!!! “ Sixth Line: It is perpendicular to 62 =+ yx at its y-intercept. Your choices are A) 32 -=- yx B) 122 =- yx C) 122 -=- yx D) 32 =- yx What do you say? You say, “All of the left sides are the same, but that’s OK, because they represent lines that are perpendicular to 62 =+ yx .” Then you say, “The y-intercept of 62 =+ yx is ( 0 , 3 ), because that point works and x=0.” Then you simply plug ( 0 , 3 ) into the equations to see where it works, and you get ….. A!!!!

18

Seventh Line: It goes through ( 5 , 3 ) and is perpendicular to 2x + 9y = 157. Your choices are……….up to you. You find it, without the help of choosing from among four. If I were you, I’d write the perpendicular line in the form Ax + By = C, because in order to be perpendicular to 2x + 9y = 157, the left side must look like this….. 9x – 2y = Now just plug in the point ( 5 , 3 ) to discover that the equation is 9x – 2y = 45-6 = 39. Eighth Line: See the picture at right. The green line is perpendicular to the red segment and passes through its midpoint. We say that the green line is the “perpendicular bisector” of the red segment. What is the equation of the green line? Oh, the coordinates of the red endpoints are ( 2 , 9 ) and ( 6 , 1 ) You say, “Good grief”, but then you notice

that the slope of the red segment is –2, so the slope of the green line is 21 . Since the midpoint

of the red segment is ( 4 , 5 ), you write the left side of the equation as 1x – 2y,

so that the slope will be 21 , and you plug in ( 4 , 5 ), to get x – 2y = -6. Done!!!

Do you remember doing a problem like that on a recent assignment? It had an x-intercept of ( - ½ , 0 ). Ninth Line: It goes through the point ( 5 , 7 ) and the x-intercept of 2x + 3y = 18.

A) 947

+-= xy B) ( )5477 -=- xy C) 6047 =+ yx D) ( )1

4714 --=- xy

It’s choice D, the only one that contains both ( 5 , 7 ) and ( 9 , 0 )

Tenth Line: It goes through ÷øö

çèæ 75.0,47 and ÷

øö

çèæ -

43,

41

A) 1044 =+ yx B) 244 -=+ yx C) 444 =- yx D) ÷øö

çèæ +=+

___

33.1822

37 xy

It’s C…………and D. Both equations are for the same line. Eleventh Line: It goes through the x-intercept of 2x+3y=14

and the y-intercept of y=7

A) 14=+ yx B) C) D) xy11107 =-

I hear you…why don’t I just make a single answer the correct one? I don’t know…but it’s A,B,C …all lines that contain ( 14 , 0 ) and ( 0 , 14 ) . OK, that’s enough.

10

8

6

4

2

-2

-4

-5 5 10 15

19

Ninth Honors Name _____________________________ 25th Assignment Period _______

1. Which line at right has a slope of 25 ? A) 652 =+ yx B) 625 =- yx C) 625 =+ yx

3. Which line at right has a negative slope? A) 832 =+ yx B) 832 =- yx C) 734 =- xy

4. Which line contains the point ÷øö

çèæ 4,21 ? A) 5x – 10y = -19 B) 1434 =+ yx C) 5

51

-= xy

5. Which lines are parallel to 543

-= xy ? A) 543

---

= xy B) )1(453 +=- yx C) 3x - 4y = 7

6. Which line is perpendicular to 832 =+ yx A) 932 =+ yx B) 832 =- yx C) 823 =- yx 7. Match each line below with its perpendicular partner. __&__ , __&__ , __&__ , __&__ , __&__ , __&__ a) 841 =+ yx b) 841 =- yx c) 814 =+ yx d) 814 =- yx e) 873 =+ yx f) 825 =- yx g) 873 =- yx h) 852 =- yx i) 825 =+ yx j) 837 =- yx k) 852 =+ yx l) 837 =+ yx Write an equation in Standard Form for each line described below. The graph paper is just for a security blanket.

8. A line whose slope is 43 and that contains ( 8 , 5 ). _____________

9. A line whose slope is 34

- and that contains ( 8 , 5 ). _____________

10. A line that has the same y-intercept as 2x + 3y = 6

but is perpendicular to it. ______________________

11. A line that contains the points ÷øö

çèæ

315,1 and ÷

øö

çèæ

314,3 ______________

12. A line perpendicular to 825 =+ yx at its y-intercept. ______________

13. A line perpendicular to 91115

--= xy and passing through the point ( 8 , 5 ) _______________

14. A line perpendicular to the segment joining ( 3 , 8 ) and ( 7 , 2 ) and passing through their midpoint. This line is called the “perpendicular bisector of the segment joining those points”.

_______________

15. The two lines that cross 797

+-= xy perpendicularly at its x and y intercepts. _________________

_________________

8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

20

9th Honors Math Name _____________________________ 26th Assignment Period _______

Write an equation in Standard Form for each line described below.

1. A line whose slope is 43

and that contains ( 1 , 10 ). 1. _______________________

2. A line parallel to 154

-= xy that has the same y-intercept as 3x + 5y = 45 2. _______________________

3. A line perpendicular to 732125

--= xy and passing through the point ( 10 , 6 ) 3. _______________________

4. A line perpendicular to 1525 =+ yx at its x-intercept. 4. _______________________

5. A line that contains the points ÷øö

çèæ

527,1 and ÷

øö

çèæ

535,4 5. _______________________

6. A line perpendicular to the segment joining ( 3 , -10 ) and ( 9 , 2 ) and 6. ________________________ passing through their midpoint.

7. The perpendicular bisector of the quadrant 1 segment of 2x + 6y = 24. 7. ________________________

8. Fill in the chart at right

Let A=( 7 , 2 ) B = ( -1 , 6 ) and C = ( -3 , -4 ) . Graph paper on back if needed.

9. Find the x-intercept of the line through C perpendicular to segment AB. 9. _____ 10. Find the length of the segment joining C to the midpoint of AB. 10. ______ 11. Find the x-intercept of the perpendicular bisector of segment AB 11. _____ 12. Find the y-intercept of the perpendicular bisector of segment BC 12. _____ 13. Find the x-intercept of the line joining the midpoints of segments AB and AC 13. _____ 14. What’s the approximate distance (to the nearest tenth) between each pair of points below?

÷øö

çèæ

53,

51

and ( )0.1 , 1.0- ______ ÷øö

çèæ 1,52

and ÷øö

çèæ

59,

56

______ ÷øö

çèæ

3029,6.0

__ and ÷

øö

çèæ

61,

151

______

15. What is the slope of the line whose equation is yxxxyx +-+=-++ 7728472 ? _____ 16. Solve the equations ( ) 87)4(35 2 +-=-+- xxxxx _______ and )73(2)43(2)1(20 +=--- xxx _____ 17. What is the area of the circle whose equation is ( ) )5()5(5)6(44 ---=-+- yyyxxx ? _______

x-intercept

y-intercept

Mdpt of intercepts

the slope

Slope- Intercept Form Equation

Standard Form Equation

( -10, 0 )

( 0 , -5 )

(15 , 0 )

53

-

2x – 3y = 12

( 12 , 0 )

( 6 , -4 )

y=x+4

( 0 , -9 )

0

2

52

-= xy

21

Notes for the 27th Assignment

Solving Linear Systems I forgot to put graph paper on the back of that last assignment…my bad. However, I was intrigued at how many of you were able to do those problems without using any graph paper…I salute you for your environmentally aware stance…saving on trees by not using graph paper.

While we’re talking about saving graph paper, I know a place where 99.2% of you would immediately

reach for one or two sheets… I give you equations of two lines and ask you the coordinates of their intersection. I know you CAN find those coordinates by graphing, unless they are something like ( 2008 , -1475 ) or

unless they are something like . How would you be able to read the coordinates that exactly?

Well, I know a way to handle both situations described above, and not need graph paper, and be faster than graph paper would have been. Here’s the idea.

Consider the graph at right, of two lines. It appears that they intersect at a point that does not have integer coordinates. In fact, the x-coordinate looks like about one and a half, and the second coordinate looks like a little over 2. Their equations are, in slope intercept form, y = 3x – 2 and y = - 2x + 5

Let’s see if our preliminary guess, that x is , is correct.

When I put in for x in the equation y = 3x – 2, I get y = 3 ( ) - 2, which makes y =

When I put in for x in the equation y = - 2x + 5, I get y= -2 ( ) + 5, which makes y = 2.

These are not the same, but they are close. What we need is an x value that will make both the blue and the red y values be the same. That is to say, We want, 3x – 2 and - 2x + 5 to give us equal values, and to find out the value of x that does this. We’ve done that before!!! We’re just saying we want to solve this equation. 3x – 2 = - 2x + 5 Recall that we just add 2x to each side, to collect all of our x’s on one side, to get 5x – 2 = 5 and that we also add 2 to each side, to collect all of our numbers on the other side, and we get 5x = 7 from which we determine, by dividing each side by 5, that x = 7/5, which is 1.4

Notice that 1.4 is almost . We guessed our x-coordinate was about , now we know that it is really 1.4 .

Now we just want to know what’s the y-coordinate that completes the ordered pair ( 1.4 , ? ). I have a simple solution, three words long, for that problem. Yep, plug it in. Plug in what? The x-coordinate we just found.

22

The blue equation gives y = 3 ( ) - 2 = 2.2 and the red equation gives y= -2 ( ) + 5 = 2.2 Notice that we get the same answer both times, which means that the point ( 1.4 , 2.2 ) is on both lines, which means that it is at the intersection of the two lines. Further notice that 2.2 is just a little larger than 2, which we predicted. Let’s try another example, this time without seeing the graphs of the lines. The lines have equations y=-4x + 11 and y = 2x – 3. Where do they intersect? Set the expressions equal. -4x + 11 = 2x – 3 then add 4x to each side, to get 11 = 6x – 3 then add 3 to each side, to get 14 = 6x

and divide each side by 6 (and reduce) . So, the intersection point is and we can find the

second coordinate by plugging in for x in the equations y = 2 – 3 = . Hence the point is .

Ok, a third example, this time with the lines whose equations are and .

What are the steps? First, set the expressions equal to each other…… =

If the presence of fractions is a little intimidating, just multiply each side of the equation by a big

number before you do anything else. That has the potential to remove all fractions, if you choose a clever big number.

Suppose you multiply by 4. Then = becomes 2x + 12 = -3x + 32 ,

from which we get 5x = 20, and x = 4. Don’t forget, you want the ordered pair, and you only have ( 4 , ? ) so far. Plug 4 into an equation to get

and check to see that your other equation also gives you that same result.

What if one or both of the lines is written in standard form or pt-slope form instead of slope-intercept form? Well, we could change them both into slope-intercept form first, but I have a thought to make that part quicker, as well.

I want you to notice some things about the point ( 2 , 1 ).

1. It’s an answer to the equation 5x + 3y = 13 , because 5 ( 2 ) + 3 ( 1 ) = 13 2. It’s an answer to the equation 10x + 6y = 26 , for a similar reason. 3. It’s an answer to the equation 50x + 30y = 130 4. It’s an answer to any equation that is just 5x + 3y = 13 with each side multiplied by something. 5. It’s an answer to the totally different equation x + 6y = 8 6. It’s an answer to any equation that is just x + 6y = 8 with each side multiplied by something.

Now, behold, I tell you a mystery….

23

7. If you take an equation that ( 2 , 1 ) is an answer to and another equation that ( 2 , 1 ) is an answer to and add them, in this case 5x + 3y = 13 ____x + 6y = 8__ 6x + 9y = 21 you get an equation ( 2 , 1 ) is an answer to.

8. If you take two equations that ( 2 , 1 ) is an answer to, and subtract them in this case 5x + 3y = 13 ____x + 6y = 8__

4x – 3y = 5 you get an equation ( 2 , 1 ) is an answer to. 9. Let’s choose 10x + 6y = 26 and subtract them. x + 6y = 8 9x = 18 HEY!!!! That’s just the vertical line x = 2. If we didn’t know that the answer was ( 2 , 1 ) ,we would at least know now that the answer was ( 2 , ___ ), and it wouldn’t take long to find out what the second coordinate was. Hmmm…Let’s pick two equations that we don’t know the intersection point for…

5x + 2y = 10 3x + 2y = 4 Whatever the intersection point is… if you add or subtract the equations, you’ll get another line with that same intersection point…which should we do? Yep, subtract them, to get 2x = 6…so ( 3 , ___ ) is the

intersection point and if you put in x=3, you get 2y = -5 , so the point is .

How do you know if you should add or subtract? Hmmm, I think you should do whatever it takes to make one variable disappear. That means…if the equations are

7x – 4y = 10 3x + 4y = 5

then we should add them to get 10x = 15, from which we know the point is and by plugging in x = 3/2

we can get that y = 1/8 , so the point is .

Some of you (Patty, I expect you are one of them) are bound to be saying… “Hey, you’re always choosing equations where the coefficients are the same, so that adding or subtracting always wipes out a variable”. Yep.. but let’s try something new. What about the system 5x + 3y = 13 x + 6y = 8 Recall that 5x + 3y = 13 is the same line as 10x + 6y = 26, so just make that change, and then subtract them. 10x + 6y = 26 __x + 6y = 8 9x = 18 Oh yeah……..we did that one.

24

Let’s try another. 2x - y = 4 x + 6y = 15 Just multiply the top equation by 6 to create the system 12x - 6y = 24 x + 6y = 15 13x = 39 ( 3 , ___ ) and plug in to get ( 3 , 2 ). OR……..hey, we could have just multiplied the bottom equation instead, by two, to get 2x - y = 4 2x + 12y = 30 and then subtracted to get -13y = -26 ( ___ , -2 ) and then plugged in. Bottom line……..IF you have two equations in standard form, although you may solve for y and use the method called Substitution, you can also just add or subtract them, first multiplying one or the other (hmmm, or both?) to make sure that when you add or subtract that one of the variables disappears…then you just have a cakewalk (and a small cakewalk in fact) to get one of the coordinates. We call this method of solving systems …………………. Linear Combination, and its modus operandi is ……………………… “add or subtract multiples of the

equations to make a variable disappear”. Here’s a tough one….. solve the system 3x - 2y = 7 5x + 3y = 2 Two ways to go Make y’s go away 9x – 6y = 21 or make x’s go away 15x – 10y = 35 10x + 6y = 4 15x + 9y = 6 adding subtracting

and you either get 19x = 25, so that or you get -19y = 29, so that

Neither will be fun to substitute back in to get the other….hey, an idea!! Do both, and be done .

25

Ninth Honors Math Name ______________________________ 27th Assignment Period _____

1. Find the point of intersection of each pair of lines listed below, without using graph paper.

a) y = 2 x + 5 and y = 4 x + 8 b) y = 7 x + 2 and y = 3 x + 11

c) y = x and y = 4x – 4 d) and

2. Write an equation in slope-intercept form and also in standard form for each line graphed at right.

A) ____________________ _____________________ B) ____________________ _____________________ C) ____________________ _____________________ D) ____________________ _____________________ E) ____________________ _____________________ F) ____________________ _____________________

3. Determine the coordinates of the point of intersection of lines A and C from question 2 above.

4. Determine the coordinates of the point of intersection of lines

a) D and F from question 2 above. b) B and E from question 2 above 5. Write each linear equation described below in point-slope, slope-intercept, and standard form.

a) containing ( 3 , 5 ) with slope 2/3 b) containing ( 4 , -2 ) perpendicular to y=x

c) containing ( 50 , 10 0 ) and ( 24 , 48 ) d) perpendicular bisector of the segment

joining ( 2 , 6 ) and ( -4 , 10 ).

6. By estimating their graphs, predict the quadrant in which the intersection will occur, for each system below.

a) b) c) d)

_______ _______ _____ _______

26

7. Solve each system below.

a) y = 8 and x = -3 b) c)

8. Write a system of linear equations, in whichever styles you wish, ____________________ whose intersection is ( 10 , 3 ).

____________________ 9. Circle each equation below that has the same graph as 2x + 3y = 1.

a) 4x + 6y = 2 b) x + 1.5y = ½ c) 5xy = 1 d) 20x + 30y = 10 e)

10. Solve each system of equations below.

a) 3x + 5y = 11 b) 5x – 3y = 6 c) x + y = 7 d) x + 3y = 7 2x +5y = 9 2x + 3y = 8 x – y = 3 x + y = 11

11. Graph each system below and also give the coordinates of their point of intersection.

a)

b) x + y = 2 2x – 3y = 4

27

Notes for 28th Assignment

Linear Story Problems I’d like to show you some story problems that boil down to systems of linear equations. 1. Five pounds of chocolate and four pounds of grapes cost 48 dollars. Two pounds of chocolate and six pounds of grapes cost 39 dollars. How much does three pounds of chocolate and one pound of grapes cost? Here’s the game plan. Let c equal the cost of one pound of chocolate, and g equal the cost of one pound of grapes. Then we can say that 5c represents the cost of five pounds of chocolates, and the first sentence can be written as 5c + 4g = 48. The second sentence can be written as 2c + 6g = 39. Now we just solve the system. 5c + 4g = 48 multiply each side by 3 15c + 12g = 142 2c + 6g = 39 multiply each side by 2 4c + 12g = 78 11c = 66 c = 6 which means that one pound of chocolate costs 6 dollars Since 2c + 6g = 39 , we know that 2(6) + 6g = 39

12 + 6g = 39 6g = 27 g = 4.5 which means that one pound of grapes costs $4.50 So, how much is 3 pounds of chocolates and a pound of grapes? 3(6) + 1(4.5) = $22.50 2. A plane flies x mph. There’s a wind of y mph. Flying with the wind, the plane goes 580 mph. Flying against the wind, the plane only goes 510 mph. What is the wind’s speed? The speed with the wind, 580 mph, is the sum of both the plane’s speed and the wind speed. So, we can write the first equation… x + y = 580. Likewise, we have the “against the wind” equation, which is x – y = 510. Now we just solve the system. x + y = 580 x – y = 510 2y = 70 y = 35 mph…that’s the speed of the wind. Do you see why we decided to subtract the equations instead of add? We weren’t asked the speed of the plane, just the speed of the wind, so we should make the plane variable disappear, not the wind variable.

28

3. In a field of goats and ducks, there are 17 animals and 48 legs. How many goats are there? Let g represent the number of goats, and d represent the number of ducks. Our first equation is easy. It’s just g + d = 17. How to write the second equation? Well, each goat has 4 legs, so 4g represents all the legs that come from goats, and 2d represents all the legs that come from ducks…so our second equation is 4g + 2d = 48. Now we just solve the system. 4g + 2d = 48 4g + 2d = 48 g + d = 17 multiply by 2 2g + 2d = 34 2g = 14 g = 7 So, there are seven goats (and 10 ducks). I had a student a few years ago who had a nice imaginative way to think of this…he said, “It’s like we just pretend they are all ducks. With 17 animals, that’s 34 legs…but there are really 48 legs, so that makes 14 extra legs, so glue 2 more legs onto each duck, making 7 ducks turn into goats.” He’s now a software programmer for a company in Palo Alto. 4. Three tickets and two bags of popcorn cost 31 dollars. Six tickets and five bags of popcorn cost 67 dollars. How much does 1 ticket and 1 bag of popcorn cost? Let t represent the cost of one ticket, and p represent the cost of one bag of popcorn. Then we write the equations, as 3t + 2p = 31 and 6t + 5p = 67. Now we just solve the system. 6t + 5p = 67 6t + 5p = 67 3t + 2p = 31 multiply by 2 6t + 4p = 62 p = 5 which means that a bag of popcorn cost 5 dollars Substituting back into an equation we find that the cost of a ticket is 7 dollars, and since the question asked for the combined cost of a ticket and a bag of popcorn, our answer is 12 dollars. I had a student a few years ago who answered this question very quickly, in a sneaky (practically cheating) way. Here was her reasoning. Student : “I subtracted the equations” Me : “ Why would you do that?” Student : “ Because then you get 6t + 5p = 67 3t + 2p = 31 3t + 3p = 36 so…… 1t + 1p = 12…..and I’m done. Me : “Wow!” She’s now teaching at a community college in Seattle.

29

Ninth Honors Name _______________________________ 28th Assignment Period _______

1. Solve the system

!

y = 2x + 53x + 5y = 29

using substitution.

2. Solve the system

!

2x " y = 43x + y = 9

using linear combination.

3. Solve the system

!

y = 7x +15y = 5x + 9

using linear combination

4. Three apples and 5 flags cost $14. Four apples and 6 flags cost $17. What is the cost of an apple?

5. Two boxes of crackers and 3 boxes of dried fruit cost 15 dollars. Four boxes of crackers and 5 boxes of dried fruit cost 26 dollars. What is the price of a box of crackers?

6. Four bananas and 3 cheese sticks cost 1 dollar and 33 cents. Three bananas and 2 cheese sticks cost 96 cents. What is the cost of 7 bananas and 5 cheese sticks?

7. When a person who walks x mph gets on a conveyor belt that moves y mph, he walks at 12 mph when he walks in the same direction the belt is moving, and only 2 mph when he walks the other way. How fast is the conveyor belt moving?

Solve each system below, using either Substitution or Linear Combination, whichever seems easiest.

8.

!

2x " 5y = 23x + y = 20

9.

!

x " 4y = 2y = 2x "13

10.

!

x + 3y =13x + 9y = 20

11.

!

4x " 3y = 02x + 6y = 30

30

PrintFreeGraphPaper.com 31

Notes for 29th Assignment

Glossary of Angles and Parts of Circles

Lines that aren’t parallel cross each other, forming angles. There are lots of words related to angles, so let’s use today to get that vocabulary organized. Another glossary, like the first day of class. When the lines are PERPENDICULAR, the angles are RIGHT ANGLES. Otherwise, you get two equal small ACUTE angles ( ABE, smaller than a right angle) and another pair of large OBTUSE angles ( ABC,larger than a right angle). Regardless of whether they cross perpendicularly or not, opposite angles in a crossing (ABC and DBE) are called VERTICAL angles. ( I know, it’s not using the word like vertical/horizontal, but that’s the word everyone uses. The opposite angles in a crossing, are “vertical”. This means ABE and CBD are also vertical angles.

There are four common ways to measure angles……and we’ll look at three of them today.

1. What fraction of the circumference is chopped out by the angle. In this case, the angle is 1/8th of the circle.

2. What percent of the circumference is chopped out by the angle. In this case, the angle is 12.5% of the circle, which you can determine by changing the fraction into a percent, 1 divided by 8.

3. Degrees….we chop the circle into 360 little parts and the number of them sliced by the angle is the degree measure of the angle. Angle ABC, whose VERTEX is at the center of the circle, is called the CENTRAL ANGLE. In this example, angle ABC measures 45 degrees.

Measures of arcs of circles: The circle is chopped into a small arc, DE, called a MINOR ARC, whose measure is also 45 degrees. The rest of the circle forms a MAJOR ARC DE, whose measure is 360-45 = 315 degrees. We sometimes distinguish between minor and major arcs by including a third point on that major arc…for example if D were to be rotated 180 around B to create A…arc DAE would be major. The Length of an arc: The length of minor arc DE is different than its measure. To find its length, find the length of the circumference and then find the appropriate fraction or percent. In this case, the radius is 1, so the diameter is 2, so the circumference is 2π. Since the arc is 1/8 of the circle, its length is 1/8 of 2π, which equals 2π/8, which reduces to π/4. The Measure and the Area of a sector: In a similar fashion, the red section (called a sector) has a measure (45 degrees) and an area that is 1/8 of the area of the circle’s area. Since the radius is 1, the area of the circle is π, and the area of that sector is π/8. The sum of the areas in a triangle: Every triangle has 3 angles, whose degree measures add up to 180 degrees. When angles add up to 180 degrees, they are called SUPPLEMENTARY ANGLES. When two segments intersect the four angles created are either vertical angles (which are equal), or they are adjacent angles, and are supplementary. When angles add up instead to 90 degrees, they are called COMPLEMENTARY ANGLES. The two acute angles in a right triangle are complementary angles.

6

4

2

-2

- 4

- 6

-10 - 5 5 10

percent of circumference = 12.5%

fraction of circle = 1/8

m!ABC=45°

A

B C

32

Ninth Honors Name _______________________________ 29th Assignment Period _______

1. Fill in the table below involving angles measured in degrees, and in fraction/percents of a circle.

Questions 2-5 are T/F 2. An angle measuring 2/7th of the circle is acute. _____ 3. An angle measuring 17% is the supplement to an angle measuring 33% of the circle. _____ 4. The complement to an angle that is 1/10 of the circle is an angle that is 2/5 of the circle. _____ 5. An angle measuring 30% of the circle is obtuse. _____

6. The measures of angles ABD and CBE are shown at right.

a. Solve for x. _____ b. How many degrees is angle ABD? _____ c. How may degrees is angle CBE? _______ d. How many degrees is angle ABC? _____

7. The two acute angles of a right triangle measure x+7 and 4x+25 degrees, as shown.

a. Solve for x. _____ b. How many degrees is the smallest angle in the triangle? _____ c. How many degrees is the largest angle in the triangle? _____

8. If the major arc HD of a circle measures 265 degrees, what is the measure of minor arc HD? _____ 9. If the major arc HD of a circle measures x degrees, what is the measure of minor arc HD? _______ 10. If the smallest angle of a right triangle is 12 degrees, what is the measure of the other acute angle? ______ 11. If the smallest angle of a right triangle is x degrees, what is the measure of the other acute angle? ______ 12. If the complement of angle ABC is 50 degrees, what is the supplement of ABC? _____ 13. If the complement of angle ABC is x degrees, what is the supplement of ABC? __________ 14. If a circle of radius 6 has an arc length of 4π, how many degrees is the subtended angle (the angle formed

by connecting the ends of the arc to the center of the circle)? _____ 15. If an arc of length 10 subtends an angle of 60 degrees, what is the area of the circle? _____ 16. Complete the following chart involving arc measures, arc lengths, and sector areas.

Number of Degrees

Fraction (reduced) of a circle

Percent of a circle

20% 12.5% 40%

Radius Diameter Circum. Area Circle fraction Arc Measure Arc Length Sector Area 32π

16π

2π

4π

30 5π

36

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Notes for 30th Assignment

Angles We’ve talked about 3 of the four ways to measure angles…degrees, fraction of a circle, and percent of a circle. There’s one more way….slopes of the lines that form the angle. It’s easiest if we imagine that one of the lines is horizontal. Then we can describe how large the angle is by saying the slope of the non-horizontal line. For example, if that second line has a slope of 1, that corresponds to a 45 degree angle, see picture at right. We use the symbols to describe the measure of the angle formed by starting at A, going to B, and turning and going to C, so that the vertex/corner of importance is B. You have to be careful with slopes; they don’t add nicely like degrees. With degrees, if you add a 45 degree angle and another 45 degree angle You make a 90 degree angle, see picture at right. However, if you think of 45 degree angles as being associated with Slopes of 1, then you find yourself wondering if an angle whose slope Is 1…..added to another angle whose slope is also 1, should give a Slope of 2…but clearly the 90 degree angle has INFINITE slope. Slope is a very useful way to measure angles, but don’t add them like you would degrees. To measure in degrees, one typically uses a protractor. We’re going to experiment in class with them. Recall that you place the vertex (corner) of the angle at the little hole in the protractor. Then you rotate the protractor around until the mark that says zero degrees lines up with one of the sides (rays) of the angle, and the other side (ray) passes through the numbers on the circumference of the protractor. Where that second side pokes through, that number is called the measure of the angle. Sound familiar? There’s a protractor on the back of your assignment, with graph paper to help you see how slope and degrees relate. Draw a line from the little hole with a slope of 1. Do you see that it hits the protractor at the 45 degree mark? Now more experimentation. If you draw a segment through the origin (hole in the protractor) with a slope of 1/3, what angle does it make with the positive x-axis? (Try it, and see if you get about 18 degrees) Let’s draw the lines whose equations are x + 2y = 12, 3x – y = 1, and y=2. They form a triangle, and we measure the three angles in that triangle, by drawing a line of slope 3 (ML) from the little hole, and noticing that it hits at about 72 degrees. Then we draw another line with slope of ½ (NM) and notice that it hits the protractor at about 27 degrees. That leaves about 81 degrees for the remaining angle, because their sum has to be 180. We write what we’ve found in this fashion...

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9th Honors Name _____________________________ 30th Assignment Period _______

Use the protractor and graph paper on the back

1. How many degrees is the angle of inclination of a line with slope…

a) of 1? _____ b) of 2? _____ c) of ? _____ d) of 3? ____ e) of ? ______

2. How many degrees is the acute angle formed by the two lines whose slopes are given below?

a) slopes of 1 and 2? ____ b) slopes of and 2? ______ c) slopes of and 3? _____

3. How many degrees is the OBTUSE angle formed by the two lines whose slopes are given below?

a) slopes of 1 and 2? ____ b) slopes of and 2? ______ c) slopes of and 3? _____

4. How many degrees is the angle of inclination of a line with slope (give the obtuse answer)…

a) of –1 ? ____ b) of –4 ? _____ c) of – ? _____ d) of –5? ____ e) of ? _____

5. Finish the sentence…when the slopes are reciprocals of each other, the angles are ______________________ 6. Finish the sentence…when the slopes are negatives of each other, the angles are ______________________

7. What are the measures of the three angles in a 3 – 4 – 5 right triangle? _______ ______ and ______

8. What are the measures of the acute angles in a right triangle whose legs are 4 and 1? ______ and _____

9. When the midpoint E of the hypotenuse in the 4 – 8 – triangle

is connected to the right angle CAD, see picture at right, several angles are formed. What are their measures?

10. Fill in the table comparing angles (to the nearest degree) with their slopes.

Slope Angle Slope Angle Slope Angle Slope Angle 4/1 _____ 4/2 _____ 4/3 ______ 4/4 _____ 3/1 _____ 3/2 _____ 3/3 _____ 3/4 _____ 2/1 _____ 2/2 _____ 2/3 _____ 2/4 _____ 1/1 _____ 1/2 _____ 1/3 _____ 1/4 _____

11. The degree measures of two equal angles are x + 35 and 2x – 25. Are they acute or obtuse? ______

12. The degree measures of two complementary angles are x + 5 and 2x – 35. Which angle is larger? _____

13. The degree measures of two supplementary angles are 3x – 2 and 5x + 6. By how many degrees does the

larger angle exceed the smaller? _____

14. The degree measures of two vertical angles are 3x – 2 and x + 10. By how many degrees does the larger angle exceed the smaller? _____

35

100

80

110

70

120

60

130

50

140

40

150

3016020

17010

1800

9090

100

80

110

70120

60130

50140

40

150

30

16020

1800

17010

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